题目内容
若函数y=f(x)存在反函数y=f-1(x),由函数y=f(x)确定数列{an},an=f(n),由函数y=f-1(x)确定数列{bn},bn=f-1(n),则称数列{bn}是数列{an}的“反数列”.(1)若数列{bn}是函数f(x)=
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(2)若函数f(x)=2
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(3)对(2)题中的{dn},不等式
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【答案】分析:(1)由f(x)=
,知f-1(x)=2x-1,所以bn=2n-1,由此能求出Sn.
(2)由f(x)=2
,知
,由此能求出{dn}的通项公式.
(3)记
+…+
,得
,故Tn+1-Tn=
>0,由此能求出实数a的取值范围.
解答:解:(1)∵f(x)=
,
∴f-1(x)=2x-1,
所以bn=2n-1,
Sn=2(1+2+3+…+n)-n
=2×
-n=n2.(4分)
(2)∵f(x)=2
,∴
,
所以dn=
.
(3)记
+…+
,
得
,
Tn+1-Tn=
>0,
所以{Tn}递增,故(Tn)min=T1=1.
由已知得,
,
解得0<a<
,
∴实数a的取值范围是(0,
).
点评:本题考查数列与函数的综合应用,解题时要认真审题,仔细解答,注意反函数的合理运用,合理地进行等价转化.
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(2)由f(x)=2
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(3)记
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解答:解:(1)∵f(x)=
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∴f-1(x)=2x-1,
所以bn=2n-1,
Sn=2(1+2+3+…+n)-n
=2×
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(2)∵f(x)=2
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所以dn=
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(3)记
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得
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Tn+1-Tn=
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所以{Tn}递增,故(Tn)min=T1=1.
由已知得,
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解得0<a<
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∴实数a的取值范围是(0,
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点评:本题考查数列与函数的综合应用,解题时要认真审题,仔细解答,注意反函数的合理运用,合理地进行等价转化.
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